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Homework Help: Harm osc, anharmonic perturbation

  1. Jan 2, 2008 #1
    1. The problem statement, all variables and given/known data
    Particle bound by
    [tex] V = \frac{1}{2} m \omega^2 x^2 - a x^3 [/tex]
    for small x. Show that the mean position of the particle changes with the energy of the eigenstates when [tex]a[/tex] is small, so first order perturbation theory works.

    2. Relevant equations
    For the harmonic oscillator
    [tex] x = \sqrt{\frac{\hbar}{2m\omega}}(a^{\dagger}+a) [/tex]

    3. The attempt at a solution
    That x^3 perturbation will give an odd number of creation/destruction operators, so there's no shift in energy eigenvalues to first order in perturbation theory. But how does that help answering the question?
  2. jcsd
  3. Jan 2, 2008 #2

    Dr Transport

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    Gold Member

    There is a first order transition, but it is between different energy states.
  4. Jan 4, 2008 #3
    But how does that help answering the question?
  5. Jan 12, 2008 #4
    Noone have any ideas on this?
  6. Jan 22, 2008 #5
    I'm not going to give up on this! The question asks to calculate the mean position of the particle, so if I do that:
    [tex]<x> = \sqrt{\frac{\hbar}{2m\omega}}<n|a^{\dagger}+a|n> = 0 [/tex]
    That's just zero, haveing nothing to do with energy eigernstates or size of a?

    Sure I could calculate the perturbated energy to second order, getting terms like
    [tex] a^2 \big( \frac{\hbar}{2m\omega} \big)^3 <n|a^{\dagger} aa \frac{1}{E_n-H_0} a^{\dagger} a a^{\dagger} |n> = -n(n+1)^2 \frac{\hbar^2 a^2}{8m^3 \omega^4}[/tex]
    and so on, but I don't see how that has anyhting to do with the question asked.
  7. Jan 22, 2008 #6
    Shouldn't you consider the expanded eigenkets?

    [tex] | n > = |n^{(0)}>+\lambda |n^{(1)}> + \lambda^2 |n^{(2)}> + \dots[/tex]

    Then you get to first order something like:

    [tex] <x> = < n^{(0)} | x | n^{(0)} > + \lambda \left( < n^{(0)} | x | n^{(1)} > + < n^{(1)} | x | n^{(0)} > \right)+ \mathcal{O} (\lambda^2) = \lambda \left( < n^{(0)} | x | n^{(1)} > + < n^{(1)} | x | n^{(0)} > \right)+ \mathcal{O} (\lambda^2)[/tex]

    Could that be the correct approach?
    Last edited: Jan 22, 2008
  8. Jan 22, 2008 #7
    I think that you dont really have to solve for the corrected energy's. You need the new eigenkets, and then just solve for the expectation value, <x>, by taking the inner product of x and the eigenkets.
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